In this channel, I break down and explain common stopper topics students run into in their middle school and high school math classes. Have a stopper topic you need explained? Leave it in the comments, and I'll get a new video posted helping break it down ASAP. Let's get mathing!
Could you somehow extend the video slightly before posting the other video suggestions. The actual answer is covered by them (I know you can turn on subtitles or read along with the transcript)
@@helpwithmathing That was quick! Felt sure it was going to be few hours. I'm in the UK and I assumed you are in the US Just after 11 am here, so thought you would still be asleep!
y = 2x² -16x + 16 ( i put the problem into the proper quadratic form of y = ax² + bx + c.) Now use vertex form of y = a(x-h)² + k h = 1/2(16/2). h = 4 [the h term, is found with the formula 1/2(-b/a)] k = -2(4²) + 16. k = -32 + 16. k = -16 [the k term is found with the formula -a(h²) + c] h = 4 and k = -16 y = 2(x-4)² - 16 [this is vertex form what we already know at this point is the vertex [h,k] and the y intercept which is just the c term in the original equation.] x = 4 + √16 / √2 and x = 4 - √16 / √2. [the formula here is x = h + √-k divided by √a and also x = h - √-k divided by √a] x = 4 + (4√2) / 2 and x = 4 - (4√2) / 2 [we cant have an irrational denominator so we multiply the denominator and the numerator by √2. √x * √x just equals x] x is the solution set [4 + 2√2 , 4 - 2√2] [ I was able to factor out a 2 from the top and bottom since it was multiplication and division] This is a lot of work shown, but if I was to do this for my own purpose or to show to a teacher who knows what I am doing then all the extra work is just explanation and it took me maybe 10 seconds to get the answer.
this is vertex form essentially. Vertex form for a quadratic formula is y = a(x-h)² +k where h = 1/2(-b/a) and k = -a(h²) + c. the a b and c terms are the same as they are in the standard quadratic form of y= ax² + bx + c. so apply all this to the equation x² - 20x + 2 =o. We rewrite this as y = x² - 20x + 2. (if we set all the x terms to o we see the y intercept is 2. This is a bonus step) now we use our formula above: h = 1/2(20/1) this of course is 20/2 which is 10 k = -1(10²) + 2. this gives me -100 + 2 which is -98. (remember if there is no coefficient before a term then assume it is 1) This gives us y = (x-10)² - 98 now solve for x with the following formulas x = h + √(-k/a) and x = h - √(-k/a) [the ( ) are used to show that the √ applies to both the -k and a term] This gives us x = 10 + √98 and x = 10 - √98 (the a term is a 1 so we can ignore it here) √98 is not a perfect square. 49 * 2 also equals 98 and 49 is a square of 7 so the square root of 98 is 7√2 This means that the solution set of x is [10+7√2, 10-7√2] this and the quadratic formula are the two ways I learned how to solve quadratic equations. You just need it in the form of ax² + bx + c = 0 then you have enough information to convert it to vertex form. A lot of work shown above, but if I wasn't explaining each and every single thing it would of taken maybe 10 seconds to do.
This is WAY too complicated. The X-Box method can be solve so much quicker and easier and without so much guess and check. I'd be happy to show you how if you'd like.
I personally agree with you, but there are all kinds of learners, and some find the box method to be the perfect balance of understanding why they are doing what they are doing and keeping everything tidy and organized. But, I reach for the X method much more. Take a look at these videos on my channel showing the X-method and see if you are satisfied. If not, I'd love to hear a new method. Check these out: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-Q_pVMvvDbRo.html
And actually, just take a peek at my factoring playlist: I'd love to hear new ideas since people are always struggling with factoring: ru-vid.com/group/PLThFCXwjOyB9KcWVAHqY4Qvs1bvpg8wRq
@vespa2860 Nice to hear from you!! Yes, I looked at the 2 in front of the x^2 term and decided just to cut to the chase with the quadratic formula, but completing the square works too and always feels so elegant to execute. Thanks for always taking the time to write in!
Mark any point on the edge of the pizza as point 1. Draw a line from that point to a second point on the edge about 90° away and mark as point 2. (Being close to 90° is not critical.) From point 2 draw a second line exactly 90° from the first line and mark its intersection with the edge of the pizza as point 3. Draw a straight third line from point 1 to point 3 and cut along that line. Assume the box has straight edges at 90° angles to get the necessary 90° angle between the first two lines..
this demonstrates an issue that I have had - why does it have to be 0 on right hand side. Well - Now I know - if you are dealing with square -there are not mutiple roots. so you do not have to have 0 on right. but if there are multiple roots, then it has to be zero on right - so that the equation can be set correctly ( if not zero - that raises too many possible answers ). Ok, got it.
I might phrase this slightly differently: if you do not have a middle term (and x term, as opposed to an x^2 or a constant) then you can get your constant to the right side and take the square root of both sides (you will still end up with a plus or minus answer on that right side). If you have a middle term, you need to get a zero on the right side of the equation to be able to apply the rule of zero once you factor (the only way two things can multiply to zero is if one or the other of them is zero)
@@helpwithmathing Yep! It's another method to add to the toolbox. I'm always on the lookout for calculation methods on the same concept, and also different expressions of the same relation. So this 'X' method of validating quadratic roots (to complement the columnar approach) is a good advantage in encapsulating the products, addends, subtrahends, and minuends in a nice, neat little graphic. It's really neat, I like it a lot. ^-^
@gamemakingkir667. Best compliment is that you came here looking for something different and still enjoyed the video! Thanks!! BTW: if you check my factoring playlist you can find that cubic video you were searching for!!
@vespa2860. LOL, I know!! I only realized as I said it out loud in the mic that I used an absurd unit when I typed up the problem. Ah well. Perhaps imagining a 25 mile high boat on utterly calm seas will give everyone a smile in difficult times. Thanks as always for watching and giving feedback
YAAAAEEEE!!!!! Finally. Thank you so much. I have been trying to understand the logic behind this. I seem to have a hard time remembering rules that I don't understand. I'm immediately going to go back to my worksheet that has a number of equations I have not been able to figure out, and I am sure I can do it now.
Not really familiar with the box method, so I tried it on your 4x^2-17x-15=0 With some effort I managed it using the factors of 4x^2 to be 4x and x, but I was unable to do it using 2x and 2x. Should I always use always use x as one of the factors? eg. x and 6x for 6x^2 as opposed to 3x and 2x. Still working through your very good content!
Hello! You ask great questions! I also find the box method to be a little too trial and error for my taste but some students really enjoy the visual of it. In answer to your question: each problem requires a different combo of first and second terms in the factors, so you can't count on always using x as one of the factors. As you discovered in the problem you worked with, there is no solution using 2x and 2x, but let's say the middle term in that problem had been 4x instead of -17x. Then the only way to solve would be to split the 4x^2 up into 2x and 2x and pair them with -3 and +5 (2x-3)(2x+5) which becomes 4x^2+10x-6x-15 or 4x^2+4x-15 and there is no solution with a 4x and an x. So, for each problem, you need to look at it individually and assess what will make the numbers work.
@@helpwithmathing Very nice explanation In the example I used. I quickly discovered that 2x wasn't possible, but relatively easy using 4x and x. edit - So if the coefficient of the x^2 term is a prime the box method should be fine. But as the factors of the x^2 coefficient increase, so does the potential difficulty. Think I will stick to the X method and then group if necessary.
I don't know why anyone would need to factor a quadratic equation, but if the exam said factor it, then why memorize the X method when everyone should know the quadratic solution formula for solving ax^2+bx+c=0; x=[-b±√(b^2-4ac)]/2a? The X method is just a modified version of this formula. To illustrate let (2x/3-4/5)(5x/7+2/3)=(10/21)x^2-(8/63)x-8/15=0. The quadratic formula gives x=6/5 and x=-14/15. Thus, (x-6/5)(x+14/15)=0, and this expression may be multiplied by anything without changing it. The coefficient of x^2 is 10/21 and how may it be factored? (2*5)/(3*7). How about (2/3)(5/7)? [(2/3)(x-6/5)][(5/7)(x+14/15)]=(2x/3-4/5)(5x/7+2/3)=0 gives back the original factored form. This was a very complex example, so let's try a simpler one. (2x-3)(5x+4)=10x^2-7x-12=0. The quadratic formula gives roots x=3/2 and x=-4/5. Thus (x-3/2)(x+4/5)=0. Factor the leading coefficient 10 into 2*5 and distribute this as (2*5)(x-3/2)(x+4/5)=[2(x-3/2)][5(x+4/5)]=(2x-3)(5x+4)=0, which matches the original factored form. Compare this method with the X method to see the similarities.
@roger7341 Thanks for the great response. You are absolutely correct that there are many many ways to factor quadratics (and many reasons to: to find the zeros of your parabola to help with manual graphing, to simplify the numerator and denominator of rational functions to help identify zeros, holes, and vertical asymptotes, or simply because you are an Algebra I student and your teacher wants to build your brain by asking you to play with factoring equations in many manners without the use of a graphing calculator. But in terms of the essential similarities of all methods, check out this video that derives the Quadratic Formula by Completing the Square on the Standard form of a Parabola. ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-b45jSYIooVE.html
Yes, my question was incorrect (not sure I am too familiar with the terminology). The discriminant was the term I was looking for. I guessed it was something to do with the b^2-4ac bit (it includes all the terms). As to your question towards the end: If the discriminant = 0 then its square root is also 0 (plus or minus 0 is still 0). This makes it essentially redundant and can be removed from the quadratic formula leaving -b/2a This makes it sure that there is only one real, rational root. For example if you factor x^2 + 2x +1 =0 you get (x+1)(x+1) these are the same and give -1 as the root. -b/2 = -2/2 = -1 Similarly for x^2 - 2x +1 =0 gives (x-1)(x-1) giving 1 as the root. gives - (-2)/2 = 1 The graph would 'kiss' the x axis at one point (-1 or 1 in my examples).
YES!! Well done!! (and there are no incorrect questions: sorry if I phrased it in a way that made you feel like you'd done something wrong there: it is a GREAT question to ask)
I'm refreshing my memory on this stuff so I can help my nephews and I seem to remember there are cases where one of the types of division cant be used. Is that true or is my brain just creating new math nightmares 30 years latet, lol?
GREAT POINT @patrickyoungmark2005 !!! Synthetic division is devised as a short cut when the divisor is in the form (x-a). So, if your divisor is non-linear (has a leading degree of higher than 1 and/or has more than two terms) you'd want to use old-school polynomial division. If your (x-a) divisor has a co-efficient on the x term greater than 1, you could divide top and bottom by that co-efficient first to get rid of it and then proceed with synthetic division.
Thanks for alerting me to the two mis-speaks!! So glad what I wrote on the whiteboard is correct. I’ll see if I can go into the RU-vid editor and fix those or use subtitles to note the things I said incorrectly but wrote correctly!
Hi again, dumb question: at 3:04 you take the square root of both sides, to eliminate the exponent power ^2. On the right, the expression is noted as plus/minus. How come the expression on the left is also not noted as +/-?
@chrishelbling3879. Let me think on this some more but my initial reaction is: it could, but it isn't worth noting since both sides having a plus/minus is the same as one side having plus minus: if both side say plus/minus you have four combinations: ++, - -, + -, - +. The ++ and - - both wind up being + and the - + and the + - both wind up being -. So you only note it on one side because it is more simplified. I'll think on it some more though, and chime in anyone who has other thoughts/Knowledge
@@helpwithmathing it's just applying the square root property that if x squared = c, then x = + or - the square root of c. Note that c is a constant while x is a variable or variable expression.
I love your channel, just discovered today. Guessing you put all this together to help your students? They probably like your voice more than Khan Academy. You are doing a great job, hope you are happy with the results, which is hard to see.
Thanks! That means a lot to me. This is a passion project of mine, to create a series of videos to help students (and parents of students) who can't afford expensive private tutoring but still want to excel at math. I've gotten some lovely feedback, so I'm going to keep chugging away for a while making new content.
